Assume $a$,$b$,$c$ are the irrational numbers, and $\sqrt a + \sqrt b + \sqrt c$ is irrational number. Show that $\sqrt{abc}$ is irrational number. Please help me this problem, thank you for watching!
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1Counter Example: $\sqrt{2}+\sqrt{3}+\sqrt{6}$. – Donald Splutterwit Mar 12 '19 at 18:35
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1@DonaldSplutterwit No it's not, because $2,3,6$ are rational. – cansomeonehelpmeout Mar 12 '19 at 18:45
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The statement is not true. However, the counterexamples in the comments don't work. Note the requirement that $a, b,$ and $c$ (not just $\sqrt{a}$, $\sqrt{b}$, and $\sqrt{c}$) should be irrational.
For a correct counterexample, take $$ a = b = c = \sqrt[3]{4}. $$ These are irrational (this can be proven similarly to the irrationality of the square root of $2$). Then, $$ \sqrt{a} + \sqrt{b} + \sqrt{c} = \sqrt[3]{2} + \sqrt[3]{2} + \sqrt[3]{2} = 3\sqrt[3]{2}, $$ which is also irrational -- because $\sqrt[3]{2}$ is irrational (which can again be proven similarly to irrationality of the square root of $2$).
However, $$ \sqrt{abc} = \sqrt{4} = 2, $$ which is rational.
Caleb Stanford
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Hans Engler's counterexample looks good to me. $\sqrt 2$ and $2^{1/4}$ are irrational, aren't they? – TonyK Mar 12 '19 at 19:07
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If $a = 2^{1/2}$ and $b = c = 2^{1/4}$, then $\sqrt{abc} = \sqrt{2}$, which is irrational. So that isn't a counterexample. – JimmyK4542 Mar 12 '19 at 19:08
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@TonyK I think Hans meant rather that $a = 2$, $b = c = \sqrt{2}$. Either way, that counterexample doesn't work. – Caleb Stanford Mar 12 '19 at 20:25
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My counterexample referred to an earlier version of the post where the claim was that $\sqrt{a}bc$ is irrational. That has been fixed by the OP and I have deleted my comment. – Hans Engler Mar 13 '19 at 14:56
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@HansEngler Thanks for the clarification, I have removed the negative reference in my answer :) – Caleb Stanford Mar 13 '19 at 17:16